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On an integral inequality

Let 0<a<b and f:[a, b] \rightarrow \mathbb{R} be a continuous and increasing function. Prove that

\displaystyle a b \int_a^b \frac{f(x)}{x^2}\, {\rm d}x \leq \int_a^b f(x) \, {\rm d}x

Solution

We are applying Chebyshev’s inequality .

     \begin{align*} \int_{a}^{b} \frac{f(x)}{x^2} \, {\rm d}x &\leq \frac{1}{b-a} \int_{a}^{b} f(x) \, {\rm d}x \int_{a}^{b} \frac{{\rm d}x}{x^2}\\ &=\frac{1}{b-a} \left ( \frac{1}{a} - \frac{1}{b} \right ) \int_{a}^{b} f(x) \, {\rm d}x \\ &= \frac{1}{ab} \int_{a}^{b} f(x) \, {\rm d}x \end{align*}

and this finishes the exercise.

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An almost zero function

Let f:[a, b] \rightarrow \mathbb{R} be a Riemann integrable function. If f(x)=0 forall rationals of the interval [a, b] then prove that \bigintsss_a^b f(x) \, {\rm d}x =0 .

Solution

Since f is Riemann integrable this means that the set of discontinuities has a zero measure. Wherever f is continuous , it’s gonna be zero due to the rationals being dense. Thus, f is almost everywhere zero. But then its Lebesgue integral is zero and so is the Riemann integral.

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The matrix is symmetric

Let A be a square matrix and for that it holds that

A^2 =AA^{\top}

Prove that A is symmetric.

Solution

Let A be an n \times n square matrix over a field \mathbb{F} such that

(1)   \begin{equation*} A^2 =AA^{\top} \end{equation*}

Taking transposed matrices back at (1) we get that

    \begin{align*} \left ( A^2 \right )^\top = \left ( A A^\top \right )^\top &\Rightarrow \left ( A^\top \right )^2 = \left ( A^\top \right )^\top A^\top \\ &\Rightarrow \left ( A^2 \right )^2 = A A^\top \end{align*}

and thus

(2)   \begin{equation*} A^2 = \left( A^\top \right)^2 \end{equation*}

On the other hand it holds that

\left ( A A^\top - A^\top A \right )^2 = \mathbb{O}_{n \times n}

since

    \begin{align*} \left ( A A^\top - A^\top A \right )^2 &= \left ( A A^\top - A^\top A \right ) \left ( A A^\top - A^\top A \right ) \\ &=A A^\top A A^\top - A A^\top A^\top A - \\ &\quad \quad -A^\top A A A^\top + A^\top A A^\top A \\ &\overset{(2)}{=} \cancel{A A A A - A A AA} - \\ &\quad \quad -A^\top AA A^\top +A^\top A A^\top A\\ &=-A^\top AA A^\top +A^\top A A^\top A \\ &\overset{(2)}{=} \cancel{A^\top A^\top A^\top A^\top - A^\top A^\top A^\top A^\top} \\ &=\mathbb{O}_{n \times n} \end{align*}

Of course it holds that if a matrix M is symmetric or antisymmetric and M^2=\mathbb{O}_{n \times n} then M=\mathbb{O}_{n \times n}. The proof is left as an exercise to the reader.

We can safely conclude using the above observations that for our matrix A it holds that

(3)   \begin{equation*} A A^\top = A^\top A \end{equation*}

But then for the matrix A-A^\top it holds that

    \begin{align*} \left ( A - A^\top \right )^2 &= \left ( A - A^\top \right ) \left ( A - A^\top \right ) \\ &=A A - A A^\top - A^\top A + A^\top A^\top \\ &\mathop {=} \limits_{(3)}^{(2)} A A - A A^\top -A A^\top + A A \\ &\overset{(2)}{=} A A - AA - AA + AA\\ &= \mathbb{O}_{n \times n} \end{align*}

and since the matrix A- A^\top is antisymmetric we conclude that A - A^\top = \mathbb{O}_{n \times n} and thus A = A^\top . Hence the result.

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On permutation

For any permutation \sigma:\{1,2,\dots,n\}\to\{1,2,\dots,n\} define its displacement  as

\displaystyle D(\sigma)=\prod_{i=1}^n |i-\sigma(i)|

What is greater: the sum of displacements of even permutations or the sum of displacements of odd permutations? The answer may depend on n.

Solution

The sum of D(\sigma) over the even permutations minus the one over the odd permutations is the determinant of the matrix A with entries a_{i,j}=\vert i-j\vert and this determinant is known to be

\det A = (-1)^{n-1} (n-1) 2^{n-2}

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Equal determinants

Let  A, B \in \mathbb{R}^{n \times n} that are diagonizable in  \mathbb{R} . If  \det (A^2+B^2)=0 and  AB=BA   then prove that

 \det A = \det B =0

Solution

The key point here is that the matrices are simultaneously diagonisable. Thus , there exists an invertible matrix C\in\mathbb{M}_{n}(\mathbb{R}) and diagonisable matrices P\,,Q\in\mathbb{M}_{n}(\mathbb{R}) such that

A=C\,P\,C^{-1}\,\,,B=C\,Q\,C^{-1}

Hence

    \begin{align*} 0&=\det(A^2+B^2)\\&=\det(C\,P^2\,C^{-1}+C\,Q^2\,C^{-1})\\ &=\det(C\,(P^2+Q^2)\,C^{-1})\\ &=\det(P^2+Q^2)\\ &=\prod_{i=1}^{n}\left(p_{i}^2+q_{i}^2\right) \quad (1) \end{align*}

where p_{i}\,,1\leq i\leq n are the diagonial elements of P and q_{i}\,,1\leq i\leq n of Q. According to (1) we have that

p_{i}^2+q_{i}^2=0 \quad \text{for some} \; i\in\left\{1,...,n\right\}

But p_{i}\,,q_{i}\in\mathbb{R}, so p_{i}=q_{i}=0 and finally we conclude that:

\displaystyle \det(A)=\det (P)=\prod_{i=1}^{n}p_{i}=0=\prod_{i=1}^{n}q_{i}=\det(Q)=\det(B)

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